A student measures the depth of a water well with an adjustable frequency audio oscillator. 2 successive resonant frequencies are heard at 4

Question

A student measures the depth of a water well with an adjustable frequency audio oscillator. 2 successive resonant frequencies are heard at 40Hz and 50Hz. What is the depth of the well?

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Ngọc Khuê 1 year 2021-08-11T14:20:37+00:00 1 Answers 1 views 0

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    2021-08-11T14:22:24+00:00

    Answer:

    17.15 m

    Explanation:

    Assuming the well is empty (full of air instead of water), the speed of sound is v = 343 m/s.

    The water well acts as a pipe closed at one end.  Therefore, the depth of the water well must be an odd multiple of a quarter of the resonant wavelength.

    L = (2n − 1) λ/4, n = 1, 2, 3, etc.

    v = λf, so λ = v/f.  Substituting:

    L = (2n − 1) v/(4f)

    Solving for frequency:

    f = (2n − 1) v/(4L)

    The difference between two successive resonant frequencies is therefore:

    Δf = (2(n+1) − 1) v/(4L) − (2n − 1) v/(4L)

    Δf = (2n + 1) v/(4L) − (2n − 1) v/(4L)

    Δf = 2 v/(4L)

    Δf = v/(2L)

    Plugging in values:

    50 Hz − 40 Hz = 343 m/s / (2L)

    2L = 34.3 m

    L = 17.15 m

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